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a(n) = Sum_{d|n, d-1 is cube} d.
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%I #11 Aug 09 2023 00:52:52

%S 1,3,1,3,1,3,1,3,10,3,1,3,1,3,1,3,1,12,1,3,1,3,1,3,1,3,10,31,1,3,1,3,

%T 1,3,1,12,1,3,1,3,1,3,1,3,10,3,1,3,1,3,1,3,1,12,1,31,1,3,1,3,1,3,10,3,

%U 66,3,1,3,1,3,1,12,1,3,1,3,1,3,1,3,10,3,1,31,1,3

%N a(n) = Sum_{d|n, d-1 is cube} d.

%F G.f.: Sum_{k>=0} (k^3+1) * x^(k^3+1)/(1 - x^(k^3+1)).

%t a[n_] := DivisorSum[n, # &, IntegerQ[Surd[#-1, 3]] &]; Array[a, 100] (* _Amiram Eldar_, Aug 09 2023 *)

%o (PARI) a(n) = sumdiv(n, d, ispower(d-1, 3)*d);

%Y Cf. A001093, A113061, A359937, A359942, A359944.

%K nonn

%O 1,2

%A _Seiichi Manyama_, Jan 19 2023